Chapter 3.8, Problem 38E

a.

To determine

To find: an equation that represents the given situation.

Answer to Problem 38E

  $I=\frac{k}{d^2}$

Explanation of Solution

Given information:

Let the intensity of light be $I$ and the distance be $d$ .

The variable I vary inversely to $d^2$ .

Calculation:

The equation for the given condition will be as follows:

  $I=\frac{k}{d^2}$

b.

To determine

To find: the value of I (round to the nearest tenth) directed at the edge of table

Answer to Problem 38E

  $13.6lux$

Explanation of Solution

Given information:

Let the intensity of light be $I$ , the distance be $d$ and the diameter of circular table be D.

The variable I vary inversely to $d^2$ .

  $\begin{array}{l}I=16lux\\ d=6feet\\ D=5foot\end{array}$

Calculation:

The equation for the given condition will be as follows:

  $I=\frac{k}{d^2}$

Put the values of I and d in the given equation and evaluate the value of $k$ as shown below.

  $\begin{array}{c}I=\frac{k}{d^2}\\ k=I{d}^2\left[\text{Rearranged the equation}\right]\\ k=16{\left(6\right)}^2\\ =576\end{array}$

Look at the figure below to understand the given situation.

  

The square distance $\left(d_1^2\right)$ between the light source and the edge of the table can be evaluated by using the Pythagoras theorem as follows:

  $\begin{array}{c}{d}_1{}^2=d^2+{\left(\frac{D}{2}\right)}^2\\ d_1{}^2={\left(6\right)}^2+{\left(\frac{5}{2}\right)}^2\left[\text{Substitute }d\text{ and }D\right]\\ d_1{}^2=36+{\left(2.5\right)}^2\\ d_1{}^2=36+6.25\\ =42.25\end{array}$

Put the values of k and $d_1{}^2$ in the equation as follows:

  $\begin{array}{c}I=\frac{k}{{\left(d_1\right)}^2}\\ =\frac{576}{42.25}\\ =13.6lux\end{array}$

Thus, the value of I is $13.6lux$ .